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Re:几行代码就行In Reply To:Re:几行代码就行 Posted by:dang at 2009-03-21 00:01:54 > > 问你个问题，怎么证明完全平方数有奇数个相因数？ 由整数的唯一分解定理知，任意一个数p=(p1^a1)*(p2^a2)....(pi^ai) pi为素数，所以p的因子的个数为(a1+1)*(a2+1)...(ai+1)这个要为奇数那么ai就要是偶数，得证 Followed by: Post your reply here: |
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